IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

rajat

if f (x) = _{[0 ]}

^{[x ]} [dx / {f(x)^2} ]

and f(2/3) = 3^{[2 ]}

2

then find f (72)

(mind u its an objective question, so no long methods allowed )

ruhi

hi can u make dis clear..............f(2/3) is 3(2)^1/2 ?????????

rajat

of course .what is so confusing in it ? ne ways plzz. solve it

KAB

Apply Newton-Leibniz rule which states that

If h(x),p(x),q(x) and f(t) are 4 functions such that

h(x) =_[q(x)}

^{[p(x) ]} f(t).dt

then

d(h(x))/dx = f(p(x))*(d(p(x))/dx) - f(q(x))*(d(q(x))/dx)

So here h(x)=f(x) , p(x)=x , q(x)=0 and f(t)=1/(f(x))²

On applying this rule,

f '(x)=(1/(f(x))²)*1-f(0)*0=1/(f(x))²

Which implies that (f(x))²f '(x) = 1 [Correction]

Or (f(x))³=x+c

Given f(2/3) = 3^{[2 ]}

2

So (3^{[2 ]}

2)³ = 2/3+c

Find c and then find f(72)

I think this is the shortest method!

rajat

plzz. use some method which will be known to everybody . i mean ....mot many ppl. know this newton--- whatever theorem

KAB

Hey! I've explained the formula.
But what is the final answer?

rajat

dont know. tell me urs

rajat

dont know. tell me urs but try solving without it

ruhi

hi rajat..........usually such type of ques. r solved by NEWTON-LEIBNITZ onli..........n kab's soln is absolutely correct........n dis formula is quiet imp. n very useful................i dont think there is ne other method 2 do dis ques :)

truly

@ KAB....

u were right by applying N-leibniz rule, but u made an error in integration....see here

u got f '(x) = 1/(f(x))²...this is right.................eqn 1

but whn u integrate LHS f'(x) wrt x is = f(x) ,,,, but whn u integrate 1/(f(x))² wrt x it is NOT -1/f(x) ....

u shud do it like this ...whn u hav got f '(x) = 1/(f(x))²
==> (f(x))² f'(x) = 1 not this can be integrated wrt f'(x)

==> (f(x))³ / 3 = f'(x) + C now putting the value of f'(x) from eqn 1

==> (f(x))³ / 3 = 1/(f(x))² + C

use the condition f(2/3) = 3^{[2 ]}

2 to find C

and then u can calculate f(72)....

This method is complicated .....

again using eqn 1 which was f '(x) = 1/(f(x))²
== > (f(x))² f'(x) = 1 = dM (say)
integrate wrt x

M = x + c1 ......... eqn 2

also M =

(f(x))² f'(x) dx .....use

uv dx

=> M = (f(x))³ -

2f(x)f'(x).f(x) dx = (f(x))³- 2M

=> M = (f(x))³ /3 + c2 ........eqn 3

from eqn 2 & 3

(f(x))³ /3 = x + C

put x = 3/2 ,,,,find C

and thn put x = 72 to find f(72)

I hope i cleared ur doubt where u were wrong

KAB

Thanks truly bhaiya.I dont know how I commited that mistake!!!

KAB

Truly bro ,plz explain a bit more after this step (f(x))^2 f'(x) = 1
Shouldnt it be like this???
On integrating (f(x))^2 f'(x) = 1
we get ((f(x))^3)/3=x+c ???
Cant we write it directly???Should we take the substitution as M and then proceed???

rajat

exactly!! even i got this

y^3 = 3x + k

but not getting the ryt answer

ruhi

hello trulely sir...........i m bit confused.........plz help...........
in de 1 st step u integrated (f(x))^2 f'(x)............wrt f'(x)...we get
(f(x))^3/3 = f'(x) + c..........so frm here v ll get de value of c
now in de 2nd step u integrated (f(x))^2 f'(x) wrt x soo by takin m nd solvin we get another eqn.......
so my ques is.........by puttin f(2/3) in de last eqn.......v vll get de value of c n then by puttin x=72 we ll get f(72)
so wot is de need of de 1st step........m not gettin
plz explain

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